BdMO 2020 Regional Higher Secondary P1
In a triangle ABC, AB=30, BC=50 and AC=60. Length of CF = k × Length of BE. Find k?
- Anindya Biswas
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Re: BdMO 2020 Regional Higher Secondary P1
We know that area of a triangle= base×height/2
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
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— John von Neumann
Re: BdMO 2020 Regional Higher Secondary P1
I used perpendicular lemma never thought like this. sad
- Mehrab4226
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Re: BdMO 2020 Regional Higher Secondary P1
What is the perpendicular lemma?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
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- Anindya Biswas
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Re: BdMO 2020 Regional Higher Secondary P1
Let $A,B,C,D$ be four points in the plane. $AC\perp BD$ if and only if $AB^2-BC^2+CD^2-DA^2=0$. Very important lemma, whoever reading this, try to prove it.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Mehrab4226
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- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BdMO 2020 Regional Higher Secondary P1
Anindya Biswas wrote: ↑Sun Mar 21, 2021 7:47 pmLet $A,B,C,D$ be four points in the plane. $AC\perp BD$ if and only if $AB^2-BC^2+CD^2-DA^2=0$. Very important lemma, whoever reading this, try to prove it.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
Re: BdMO 2020 Regional Higher Secondary P1
That's an amazing solution. I also solved the problem but used a long naive area based solution. Thank you very much, learned a new way to think.Anindya Biswas wrote: ↑Sun Mar 21, 2021 3:46 pmWe know that area of a triangle= base×height/2
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$